Altitudes $\overline{AX}$ and $\overline{BY}$ of acute triangle $ABC$ intersect at $H$.  If $\angle BAC = 61^\circ$ and $\angle ABC = 73^\circ$, then what is $\angle CHX$?
First, we build a diagram:

[asy]
size(150); defaultpen(linewidth(0.8));
pair B = (0,0), C = (3,0), A = (1,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);
draw(A--B--C--cycle);
draw(A--P^^B--Q);
pair Z;
Z = foot(C,A,B);
draw(C--Z);
label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$X$",P,S); label("$Y$",Q,E); label("$H$",H+(0,-0.20),SW);
label("$Z$",Z,NW);
draw(rightanglemark(B,Z,H,3.5));
draw(rightanglemark(C,P,H,3.5));
draw(rightanglemark(H,Q,C,3.5));
[/asy]

Since altitudes $\overline{AX}$ and $\overline{BY}$ intersect at $H$, point $H$ is the orthocenter of $\triangle ABC$.  Therefore, the line through $C$ and $H$ is perpendicular to side $\overline{AB}$, as shown.  Therefore, we have $$\angle CHX= 90^\circ - \angle HCX = 90^\circ - \angle ZCB = \angle ZBC = \boxed{73^\circ}.$$